Bài giải :
a.
`-n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{14}{28}=0,5(mol)`
`-n_{NO}=\frac{m_{NO}}{M_{NO}}=\frac{4}{30}=\frac{2}{15}(mol)`
`⇒n_{hh}=n_{N_2}+n_{NO}=0,2+\frac{2}{15}=\frac{19}{30}(mol)`
`⇒V_{hh}(đktc)=n_{hh}.22,4=\frac{19}{30}.22,4=\frac{1064}{75}(l)`
b.
- Vì $d_{H_2O}=1(g/ml)$
Mà `V_{H_2O}=0,8(l)=800(ml)`
`⇒m_{H_2O}=800.1=800(g)`
`⇒n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{800}{18}=\frac{400}{9}(mol)`